Each side of a metal plate is illuminated by light of different wavelengths. The left side is illuminated by light with λ0 = 500 nm and the

Question

Each side of a metal plate is illuminated by light of different wavelengths. The left side is illuminated by light with λ0 = 500 nm and the right side by light of unknown λ. Two electrodes A and B provide the stopping potential for the ejected electrons. If the voltage across AB is VAB=1.2775 V, what is the unknown λ?

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Thạch Thảo 3 years 2021-08-12T13:22:36+00:00 1 Answers 5 views 0

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  1. lethu
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    2021-08-12T13:23:55+00:00
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    Answer:

    The  wavelength is  \lambda = 1029 nm

    Explanation:

    From the question we are told that

        The  wavelength of the left light is  \lambda_o = 500 nm = 500 *10^{-9} \ m

          The  voltage across A  and  B is  V_{AB } = 1.2775 \ V

    Let the stopping potential  at A  be V_A and the electric potential at B  be  V_B

    The voltage across A and B is mathematically represented as

          V_{AB} = V_A - V_B

    Now  According to Einstein’s photoelectric equation the stopping potential at A for the ejected electron from the left side  in terms of electron volt is mathematically represented as

            eV_A = \frac{h * c}{\lambda_o } - W

    Where  W is the work function of the metal

                 h is the Planck constant with values  h = 6.626 *10^{-34} \ J \cdot s

                 c  is the speed of light with value c = 3.0 *10^{8} \ m/s

    And  the stopping potential at B for the ejected electron from the right side  in terms of electron volt is mathematically represented as

              eV_B = \frac{h * c}{\lambda } - W

    So  

          eV_{AB} = eV_A - eV_B

    =>    eV_{AB} = \frac{h * c}{\lambda_o } - W - [\frac{h * c}{\lambda } - W]

    =>   eV_{AB} = \frac{h * c}{\lambda_o } - \frac{h * c}{\lambda }

    =>   \frac{h * c}{\lambda } = \frac{h * c}{\lambda_o } -eV_{AB}

    =>  \frac{1}{\lambda } =\frac{1}{\lambda_o } - \frac{ eV_{AB}}{hc}

    Where e is the charge on an electron with the value  e = 1.60 *10^{-19} \ C

    =>   \frac{1}{\lambda } = \frac{1}{500 *10^{-9} } - \frac{1.60 *10^{-19} * 1.2775}{6.626 *10^{-34} * 3.0 *10^{8}}      

    =>  \frac{1}{\lambda } = 9.717*10^{5} m^{-1}  

    =>   \lambda = 1.029 *10^{-6} \ m

    =>   \lambda = 1029 nm

         

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